A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is ` 3. 92 xx 10 ^(-3)` m. what must ve the least period of these oscillations. So that the object is not detached from the platform

Options

A. 0.1256 s
B. 0.1356 s
C. 0.1456 s
D. 0.1556 s

Answer

A

Explanation

Acceleration `(a)=Aomega^(2)`
`mg-R=ma`
For critical condition R=0
`therefore mg=ma implies mg=mAomega^(2)`
`implies omega=sqrt(g/A)=sqrt((9.8)/(3.92xx10^(-3)))=50`
`implies T=(2pi)/(omega)=(2pi)/(50)=0.1256` sec.